Problem: Find the smallest three-digit palindrome whose product with 101 is not a five-digit palindrome.
Solution: We can use the distributive property of multiplication to multiply a three-digit palindrome $aba$ (where $a$ and $b$ are digits) with 101: $$ 101 \cdot aba = (100 + 1) \cdot aba = aba00 + aba = ab(2a)ba. $$Here, the digits of the product are $a$, $b$, $2a$, $b$, and $a$, unless carrying occurs.  In fact, this product is a palindrome unless carrying occurs, and that could only happen when $2a \ge 10$.  Since we want the smallest such palindrome in which carrying occurs, we want the smallest possible value of $a$ such that $2a \ge 10$ and the smallest possible value of $b$.  This gives us $\boxed{505}$ as our answer and we see that $101 \cdot 505 = 51005$ is not a palindrome.